Analyzing Business Data in SQL
Michel Semaan
Data Scientist
DATE_TRUNC('quarter', '2018-08-13') ? '2018-07-01 00:00:00+00:00''2018-07-01 00:00:00+00:00' :: DATE ? '2018-07-01'Dates in reports
'2018-08-13', isn't very readable'2018-08-13' to 'Friday 13, August 2018'?Solution
TO_CHAR('2018-08-13', 'FMDay DD, FMMonth YYYY') ? 'Friday 13, August 2018'TO_CHAR(DATE, TEXT) ? TEXT (the formatted date string)Example: Dy ? Abbreviated day name (Mon, Tues, etc.)
TO_CHAR('2018-06-01', 'Dy') ? 'Fri'TO_CHAR('2018-06-02', 'Dy') ? 'Sat'Patterns in the format string will be replaced by what they represent in the date; other characters remain as-is
DD ? Day number (01 - 31)TO_CHAR('2018-06-01', 'Dy - DD') ? 'Fri - 01'TO_CHAR('2018-06-02', 'Dy - DD') ? 'Sat - 02'| Pattern | Description |
|---|---|
FMDay |
Full day name (Monday, Tuesday, etc.) |
MM |
Month of year (01 - 12) |
Mon |
Abbreviated month name (Jan, Feb, etc.) |
FMMonth |
Full month name (January, February, etc.) |
YY |
Last 2 digits of year (18, 19, etc.) |
YYYY |
Full 4-digit year (2018, 2019, etc.) |
Documentation: https://www.postgresql.org/docs/9.6/functions-formatting.html
Query
SELECT DISTINCT
order_date,
TO_CHAR(order_date,
'FMDay DD, FMMonth YYYY') AS format_1,
TO_CHAR(order_date,
'Dy DD Mon/YYYY') AS format_2
FROM orders
ORDER BY order_date ASC
LIMIT 3;
Result
order_date format_1 format_2
---------- ---------------------- ---------------
2018-06-01 Friday 01, June 2018 Fri 01/Jun 2018
2018-06-02 Saturday 02, June 2018 Sat 02/Jun 2018
2018-06-02 Sunday 03, June 2018 Sun 03/Jun 2018
SUM(...) OVER (...): Calculates a column's running totalSUM(registrations) OVER (ORDER BY registration_month) calculates the registrations running totalLAG(...) OVER (...): Fetches a preceding row's valueLAG(mau) OVER (ORDER BY active_month) returns the previous month's active users (MAU)RANK() OVER (...): Assigns a rank to each row based on that row's position in a sorted orderRANK() OVER (ORDER BY revenue DESC) ranks users, eateries, or months by the revenue they've generatedQuery
SELECT
user_id,
SUM(meal_price * order_quantity) AS revenue
FROM meals
JOIN orders ON meals.meal_id = orders.meal_id
GROUP BY user_id
ORDER BY revenue DESC
LIMIT 3;
Result
user_id revenue
------- -------
18 626
76 553.25
73 537
Query
WITH user_revenues AS (
SELECT
user_id,
SUM(meal_price * order_quantity) AS revenue
FROM meals
JOIN orders ON meals.meal_id = orders.meal_id
GROUP BY user_id)
SELECT
user_id,
RANK() OVER (ORDER BY revenue DESC)
AS revenue_rank
FROM user_revenues
ORDER BY revenue_rank DESC
LIMIT 3;
Result
user_id revenue_rank
------- ------------
18 1
76 2
73 3
Analyzing Business Data in SQL